How do this? 2^(3x-1)*4^(2x+3)=8^(3-x)

2^(3x-1)*4^(2x+3)=8^(3-x)

 

We can write 4ⁿ as (2²)ⁿ → 2²ⁿ  and 8ⁿ as (2³)ⁿ → 2³ⁿ

 

so 2^(3x-1)*4^(2x+3)=8^(3-x)  becomes   2^(3x-1)*2^(4x+6) = 2^(9-3x)

 

Since 2^a*2^b = 2^a+b we can further write  2^(3x-1+4x+6) = 2^(9-3x)  or  2^{(7x + 5)} = 2^{(9 - 3x)}

 

From this it is clear that 7x + 5 = 9 - 3x   so  10x = 4  or x = 4/10 → 0.4

2^(3x-1)*4^(2x+3)=8^(3-x)

 

2^(3x-1)*4^(2x+3)=2^(7x + 5) As you can see, the left hand side does not equal the right hand side.

Solve for x over the real numbers:
2^(7 x+5) = 8^(3-x)

 

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
log(2) (7 x+5) = 3 log(2) (3-x)

 

Expand out terms of the left hand side:
7 log(2) x+5 log(2) = 3 log(2) (3-x)

 

Expand out terms of the right hand side:
7 log(2) x+5 log(2) = 9 log(2)-3 log(2) x

 

Subtract 5 log(2)-3 x log(2) from both sides:
10 log(2) x = 4 log(2)

 

Divide both sides by 10 log(2):
Answer: |x = 2/5